Are you expecting someone to do your homework for you? You need to work the problem yourself so you can learn something. Make a good effort and someone will help if you get stuck and have a specific question.

Are you expecting someone to do your homework for you? You need to work the problem yourself so you can learn something. Make a good effort and someone will help if you get stuck and have a specific question.

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NO, I don't want anybody to do my assignment. I was going to post my solution, but you were too quick!

The value of differential input resistance is 6175k. Is that taken to be from the inverting input to the non inverting input or from each input to ground?

The value of differential output resistance is 416 ohms. Is that in series with the output of the op amp or from output to ground or some other connection?

I took the input resistance to be from non inverting to inverting terminals.
I took the output resistance to be in series with the output of the (internal) op amp.
I got a voltage gain close to 18. I suppose we could sim this to check.

LATER:
Checked using a second method which involved calculating resistors in series and parallel, came out with the same result: slightly over 18 but i'll wait to post the exact solution. That is with my above assumptions about the input and output impedances.
Also note that the two main resistors are R3 and R2, and if they were the only feedback resistors the gain would be 17.4 which is close to 18 more or less.

Keep in mind that if you do it this way, you have to calculate the total resistance working from output to input, then from input to output, in order to get the proper total resistances for use in a voltage divider formula.
For example, to calculate the voltage at the inverting terminal due to the op amp very output voltage we have to do:
R51=R5||R1
R851=R51+R8
R2851=R851||R2
R32861=R2851+R3
R432851=R32581||R4

and finally, the output of the circuit is:
Vout*R432851/(R432851+Ro)
where Ro is the 416 ohm output resistance of the op amp.

So you see that there are seven resistors involved in the calculation of the voltage at the inverting terminal and that is just due to the output of the very internal op amp. We would then have to calculate forward in the same manner to get the voltage at the inverting input due to the input voltage itself. The sum of these two is the total voltage at the inverting input.
You can also see that to calculate the voltages at both the inverting and non inverting terminals requires four calculations when doing it this way.
An alternative is to use nodal analysis and using a voltage controlled voltage source for the op amp.

Since the circuit is different if we take the output resistance 416 to be from the op amp output to ground, so this has to be clearly specified. The gain however will still come out close to 18 though.

I see you redid your calculation of the gain
That's much better. I got 18.37 from input generator to output.
You can adjust for Vin by knowing the ratio of the two resistors at the input, and then can calculate input resistance too if you like.

I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?

I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?

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If the problem doesn't explicitly say that Af is to be Vout/Vin rather than Vout/Vs, I would solve it for both, a triviality. That way you are guaranteed to have the answer your professor wants, plus one extra one.

R1 is outside the feedback network, and you shouldn't include it until the end. Use these expressions:

I was thinking, that maybe we should include Rs, because in question 2 and 3 we have to find the input impedance WITH feedback, and output impedance WITH feedback. Or am I getting it wrong? Why would my teacher give the value of Rs, if I don't use it anywhere?

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Hi,

If you had an amplifier sitting around and decided to test it, you would connect a wave generator to the input. The wave generator would have some series output resistance Rs. You would turn up the generator to maybe 1v peak, and measure it at the input of the amplifier which is Vin. Knowing Vin is 1v peak and the output was 18v peak would tell you that the gain of the AMPLIFIER itself was 18. If you could look inside the generator box and saw a zero output impedance generator connected to a 100 ohm resistor, then you would know the actual output of the internal generator, which would not help with calculating the gain.

However, if you know the true internal generator voltage and the true internal generator impedance then you can calculate the input resistance of the AMPLIFIER itself because that information gives you the input current in addition to the input voltage Vin. Hence, if you know that extra information then you can easily calculate the other requirement which is the input resistance of the amplifier itself where the input is at the node labeled Vin.

In short, the input is at Vin, but knowing other external things can help calculate the input resistance. So you dont actually skip Rs except for the calculation of the gain.

If the internal generator voltage is Vs and the amplfier input is Vin, and the generator Rs is 100, then the input current is:
Iin=(Vs-Vin)/100

Basically since the input is Vin and the current is Iin, then the input resistance at that level is Vin/Iin, and now you have the input resistance too.